epsilon-delta proofs use limits.
abs(f) is continuous if f is continuous, because
|f| = f if f>=0
|f| = -f is f<0
So, if the limit L exsists for f(x), it also exists for |f(x)|.
Prove if f(p) = 0 where abs(f) is continuous at p, then f is continuous at p. Note: we have only learned continuity, discontinuty, and epsilon-delta proofs (no limits)!
1 answer